class Solution:
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        while m > 0 and n > 0:
            if nums1[m-1] < nums2[n-1]:
                nums1[m-1+n] = nums2[n-1]
                n = n - 1
            else:
                nums1[m-1+n], nums1[m-1] = nums1[m-1], nums1[m-1+n]
                m = m -1
        if m == 0 and n>0:
            nums1[:n] = nums2[:n]
        return nums1

if __name__ == "__main__":
    nums1 = [1, 3, 5, 0, 0, 0, 0]
    nums2 = [3, 4, 6, 7]
    print(Solution().merge(nums1, 3, nums2, 4))






    """
            Time Complexity = O(n)
            Space Complexity = O(1)

            Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
            Note:
            The number of elements initialized in nums1 and nums2 are m and n respectively.
            You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional
            elements from nums2.

            Example:
            Input:
            nums1 = [1,2,3,0,0,0], m = 3
            nums2 = [2,5,6],       n = 3
            Output: [1,2,2,3,5,6]

            Answer may have some problems.
    """
